POJ 3678 Katu Puzzle （2-sat,4级）

D - Katu Puzzle

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description: System Crawler (2013-05-30)

Description

Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex V_i a value X_i (0 ≤ X_i ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

 X_a op X_b = c

The calculating rules are:

AND 0 1 0 0 0 1 0 1

OR 0 1 0 0 1 1 1 1

XOR 0 1 0 0 1 1 1 0

Given a Katu Puzzle, your task is to determine whether it is solvable.

Input

The first line contains two integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.
The following M lines contain three integers a (0 ≤ a < N), b(0 ≤ b < N), c and an operator op each, describing the edges.

Output

Output a line containing "YES" or "NO".

Sample Input

4 4
0 1 1 AND
1 2 1 OR
3 2 0 AND
3 0 0 XOR

Sample Output

YES

Hint

X ₀ = 1, X ₁ = 1, X ₂ = 0, X ₃ = 1.

哎，又是一几乎模板，就是要搞清楚逻辑关系。

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=2010;
class Edge
{
  public:int v,next;
};
class TWO_SAT
{
public:
  int dfn[mm],e_to[mm],stack[mm];
  Edge e[mm*mm*2];
  int edge,head[mm],top,dfs_clock,bcc;
  void clear()
  {
    edge=0;clr(head,-1);
  }
  void add(int u,int v)
  {
    e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
  }
  void add_my(int x,int xval,int y,int yval)
  {
    x=x+x+xval;y=y+y+yval;
    add(x,y);
  }
  void add_clause(int x,int xval,int y,int yval)
  {///x or y
    x=x+x+xval;
    y=y+y+yval;
    add(x^1,y);add(y^1,x);
  }
  void add_con(int x,int xval)//x is xval
  { x=x+x+xval;
    add(x^1,x);
  }
  int tarjan(int u)
  {
    int lowu,lowv;
    lowu=dfn[u]=++dfs_clock;
    int v; stack[top++]=u;
    for(int i=head[u];~i;i=e[i].next)
    {
      v=e[i].v;
      if(!dfn[v])
      {
        lowv=tarjan(v);
        lowu=min(lowv,lowu);
      }
      else if(e_to[v]==-1)//in stack
        lowu=min(lowu,dfn[v]);
    }
    if(dfn[u]==lowu)
    {
      ++bcc;
      do{
        v=stack[--top];
        e_to[v]=bcc;
      }while(v!=u);
    }
    return lowu;
  }
  bool find_bcc(int n)
  { clr(e_to,-1);
    clr(dfn,0);
    bcc=dfs_clock=top=0;
    FOR(i,0,2*n-1)
    if(!dfn[i])
      tarjan(i);
    for(int i=0;i<2*n;i+=2)
      if(e_to[i]==e_to[i^1])return 0;
    return 1;
  }
}two;
int n,m;
char s[44];
int main()
{ int a,b,c,d;
  while(~scanf("%d%d",&n,&m))
  {
    two.clear();
    FOR(i,1,m)
    {
      scanf("%d%d%d%s",&a,&b,&c,s);
      if(s[0]=='A')
      {
        if(c)
        {
          two.add_clause(a,1,b,1);
          two.add_clause(a,0,b,1);//imposible a 1 b 0
          two.add_clause(a,1,b,0);
//          two.add_my(a,0,b,1);
//          two.add_my(a,0,b,0);
//          two.add_my(b,0,a,1);
//          two.add_my(b,0,a,0);
        }
        else
        {
          two.add_clause(a,0,b,0);
        }
      }
      else if(s[0]=='O')
      {
        if(c)
        {
          two.add_clause(a,1,b,1);
        }
        else
        {   two.add_clause(a,0,b,0);
            two.add_clause(a,1,b,0);
            two.add_clause(a,0,b,1);
//          two.add_my(a,1,b,0);
//          two.add_my(a,1,b,1);
//          two.add_my(b,1,a,0);
//          two.add_my(b,1,a,1);
        }
      }
      else if(s[0]=='X')
      {
        if(c)
        {
          two.add_clause(a,1,b,1);
          two.add_clause(a,0,b,0);
        }
        else
        {
          two.add_clause(a,1,b,0);
          two.add_clause(a,0,b,1);
        }
      }
    }
    if(two.find_bcc(n*2))printf("YES\n");
    else printf("NO\n");
  }
  return 0;
}